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Operations Research Assistant Professor Dr. Sana’a Wafa Al-Sayegh 2 nd Semester 2008-2009 ITGD4207 University of Palestine

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  • Operations ResearchAssistant Professor Dr. Sanaa Wafa Al-Sayegh2nd Semester 2008-2009ITGD4207University of Palestine

  • ITGD4207 Operations Research

    Chapter 4

    Linear Programming Duality and Sensitivity analysis

  • Linear Programming Duality and Sensitivity analysis

    Dual Problem of an LPP Definition of the dual problem Examples (1,2 and 3)Sensitivity AnalysisExamples (1 and 2)

  • Given a LPP (called the primal problem), we shall associate another LPP called the dual problem of the original (primal) problem. We shall see that the Optimal values of the primal and dual are the same provided both have finite feasible solutions. This topic is further used to develop another method of solving LPPs and is also used in the sensitivity (or post-optimal) analysis.Dual Problem of an LPP

  • Definition of the dual problemGiven the primal problem (in standard form)Maximizesubject to

  • the dual problem is the LPPMinimizesubject to

  • If the primal problem (in standard form) isMinimizesubject to

  • Then the dual problem is the LPPMaximizesubject to

  • 1. In the dual, there are as many (decision) variables as there are constraints in the primal. We usually say yi is the dual variable associated with the ith constraint of the primal.2. There are as many constraints in the dual as there are variables in the primal.We thus note the following:

  • 3. If the primal is maximization then the dual is minimization and all constraints are If the primal is minimization then the dual is maximization and all constraints are In the primal, all variables are 0 while in the dual all the variables are unrestricted in sign.

  • 5. The objective function coefficients cj of the primal are the RHS constants of the dual constraints.6. The RHS constants bi of the primal constraints are the objective function coefficients of the dual.7. The coefficient matrix of the constraints of the dual is the transpose of the coefficient matrix of the constraints of the primal.

  • Example1Write the dual of the LPPsubject toMaximize

  • Thus the primal in the standard form is:subject toMaximize

  • Hence the dual is: subject toMinimize

  • Example2Write the dual of the LPP subject toMinimize

  • Thus the primal in the standard form is:subject toMinimize

  • Hence the dual is: Maximizesubject to

  • Example3Write the dual of the LPPsubject toMaximize

  • Maximize subject toThus the primal in the standard form is:

  • Minimizesubject toHence the dual is:

  • Theorem: The dual of the dual is the primal (original problem).Proof. Consider the primal problem (in standard form)Maximizesubject to

  • Sensitivity Analysis

    Sensitivity analysis is the study of how changes

    in the coefficients of a linear program affect the optimal solutionor in the value of right hand sides of the problem affect the optimal solution

  • Sensitivity Analysis contdUsing sensitivity analysis we can answer questions such as:

    1.How will a change in a coefficient of the objective function affect the optimal solution?

    We can define a range of optimality for each objective function coefficient by changing the objective function coefficients, one at a time

  • Sensitivity Analysis contd2.How will a change in the right-hand-side value for a constraint affect the optimal solution?The feasible region may change when RHSs (one at a time) are changed and perhaps cause a change in the optimal solution to the problem

  • Sensitivity Analysis contd3.How much value is added/reduced to the objective function if I have a larger/smaller quantity of a scarce resource?Sensitivity analysis is important to the manager who must operate in a dynamic environment with imprecise estimates of the coefficients

  • Example 1LP Formulation Max 5x1 + 7x2

    s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8

    x1, x2 > 0

  • Example 1Graphical Solution

    8

    7

    6

    5

    4

    3

    2

    1

    1 2 3 4 5 6 7 8 9 10 2x1 + 3x2 < 19 x2x1x1 + x2 < 8Max 5x1 + 7x2x1 < 6Optimal: x1 = 5, x2 = 3, z = 46

  • Objective Function CoefficientsLet us consider how changes in the objective function coefficients might affect the optimal solution.The range of optimality for each coefficient provides the range of values over which the current solution will remain optimal.Managers should focus on those objective coefficients that have a narrow range of optimality and coefficients near the endpoints of the range.

  • Example 1Changing Slope of Objective Function

    8

    7

    6

    5

    4

    3

    2

    1

    1 2 3 4 5 6 7 8 9 10 x1FeasibleRegion12345 x2

  • Range of OptimalityGraphically, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines.The slope of an objective function line, Max c1x1 + c2x2, is -c1/c2, and the slope of a constraint, a1x1 + a2x2 = b, is -a1/a2.

  • Right-Hand SidesLet us consider how a change in the right-hand side for a constraint might affect the feasible region and perhaps cause a change in the optimal solution.The improvement in the value of the optimal solution per unit increase in the right-hand side is called the dual price.The range of feasibility is the range over which the dual price is applicable.As the RHS increases, other constraints will become binding and limit the change in the value of the objective function.

  • Example 2Consider the following linear program:

    Min 6x1 + 9x2 ($ cost) s.t. x1 + 2x2 < 8 10x1 + 7.5x2 > 30 x2 > 2 x1, x2 > 0

  • Example 2Optimal Solution

    According to the output: x1 = 1.5 x2 = 2.0Objective function value = 27.00

  • Example 2Range of OptimalityQuestionSuppose the unit cost of x1 is decreased to $4. Is the current solution still optimal? What is the value of the objective function when this unit cost is decreased to $4?

  • Example 2Range of OptimalityAnswerThe output states that the solution remains optimal as long as the objective function coefficient of x1 is between 0 and 12. Because 4 is within this range, the optimal solution will not change. However, the optimal total cost will be affected: 6x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00.

  • Example 2Range of OptimalityQuestionHow much can the unit cost of x2 be decreased without concern for the optimal solution changing?

  • Example 2Range of OptimalityAnswerThe output states that the solution remains optimal as long as the objective function coefficient of x2 does not fall below 4.5.

  • Example 2Range of FeasibilityAnswerA dual price represents the improvement in the objective function value per unit increase in the right-hand side. A negative dual price indicates a negative improvement in the objective, which in this problem means an increase in total cost because we're minimizing. Since the right-hand side remains within the range of feasibility, there is no change in the optimal solution. However, the objective function value increases by $4.50.